Problem Study : Geometric Vector

The diagram shows triangle $ \displaystyle OAB$ in which $ \displaystyle \overrightarrow{{OA}}=\vec{a}$ and $ \displaystyle \overrightarrow{{OB}}=\vec{b}$. The point $ \displaystyle C$ is the midpoint of $ \displaystyle OA$ and the point $ \displaystyle D$ is the midpoint of $ \displaystyle DC$.

(a) Express $ \overrightarrow{{OD}}$ in terms of $ \vec{a}$ and $ \vec{b}$.

(b) If the point E lies on AB, then show that $ \overrightarrow{{OE}}$ can be written in the form $ \displaystyle \vec{a}+k\left( {\vec{b}-\vec{a}} \right)$, where k is a constant.

      Given also that $ \displaystyle OB$ produced meets $ \displaystyle AB$ at $ \displaystyle E$,

(c) find $ \overrightarrow{{OE}}$,

(d) $ \displaystyle AB:EB=2:1$ 

Solution

$ \displaystyle \text{Since}$ $ \displaystyle D$ $ \displaystyle \text{is the midpoint of}$ $ \displaystyle BC$, $ \displaystyle \text{by midpoint formula,}$

$ \displaystyle \begin{array}{l}\overrightarrow{{OD}}=\frac{1}{2}\left( {\overrightarrow{{OC}}+\overrightarrow{{OB}}} \right)\\\\\ \ \ \ \ \ =\frac{1}{2}\left( {\frac{1}{2}\vec{a}+\vec{b}} \right)\\\\\ \ \ \ \ \ =\frac{1}{4}\vec{a}+\frac{1}{2}\vec{b}\end{array}$

$ \displaystyle \text{Let}$ $ \displaystyle E$ $ \displaystyle \text{be a point on}$ $ \displaystyle AB$ $ \displaystyle \text{such that}$ $ \displaystyle \overrightarrow{{AE}}=k\overrightarrow{{AB}}.$ 

$ \displaystyle \begin{array}{l}\therefore \overrightarrow{{OE}}=\overrightarrow{{OA}}+\overrightarrow{{AE}}\\\\\ \ \ \ \ \ \ \ =\overrightarrow{{OA}}+k\overrightarrow{{AB}}\\\\\ \ \ \ \ \ \ \ =\overrightarrow{{OA}}+k\left( {\overrightarrow{{OB}}-\overrightarrow{{OA}}} \right)\\\\\ \ \ \ \ \ \ \ =\vec{a}+k\left( {\vec{b}-\vec{a}} \right)\end{array}$

$ \displaystyle \text{Since}$ $ \displaystyle OD$ $ \displaystyle \text{produced meets}$ $ \displaystyle AB$ $ \displaystyle \text{at}$ $ \displaystyle E,$ $ \displaystyle \text{let}$ $ \displaystyle \overrightarrow{{OE}}=h\overrightarrow{{OD}}.$ 

$ \displaystyle \therefore \left( {1-k} \right)\vec{a}+k\vec{b}=\frac{h}{4}\vec{a}+\frac{h}{2}\vec{b}$

$ \displaystyle \text{Since}$ $ \displaystyle \vec{a}$ $ \displaystyle \text{and}$ $ \displaystyle \vec{b}$ $ \displaystyle \text{are not parallel and}$ $ \displaystyle \vec{a}\ne \vec{0},\vec{b}\ne\vec{0},$ 

$ \displaystyle \begin{array}{l}\ \ \ k=\frac{h}{2}\Rightarrow h=2k\\\\\ \ \ 1-k=\frac{h}{4}\Rightarrow h=4-4k\\\\\therefore 2k=4-4k\Rightarrow k=\frac{2}{3}\end{array}$ 

$ \displaystyle \text{Since }\overrightarrow{{AE}}=k\overrightarrow{{AB}},$

$ \displaystyle \begin{array}{l}\overrightarrow{{AE}}=\frac{2}{3}\overrightarrow{{AB}}\Rightarrow AE=\frac{2}{3}AB\\\\\therefore \overrightarrow{{EB}}=\frac{1}{3}\overrightarrow{{AB}}\Rightarrow EB=\frac{1}{3}AB\\\\\therefore AE:EB=\frac{2}{3}AB:\frac{1}{3}AB\\\\\therefore AE:EB=2:1\end{array}$